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<p><dfn class="terminology">Proof:</dfn> We need to show that any solution of (<a href="" class="xref" data-knowl="./knowl/eq3_11.html" title="Equation 3.6.1">(3.6.1)</a>) is included in (<a href="" class="xref" data-knowl="./knowl/eq3_14.html" title="Equation 3.6.4">(3.6.4)</a>). Consider any solution <span class="process-math">\(Y_1(x)\text{,}\)</span> we know <span class="process-math">\(Y_1(x)-Y(x)\)</span> is a solution of (<a href="" class="xref" data-knowl="./knowl/eq3_12.html" title="Equation 3.6.2">(3.6.2)</a>). As <span class="process-math">\(C_1 y_1+C_2 y_2\)</span> is the general solution of (<a href="" class="xref" data-knowl="./knowl/eq3_12.html" title="Equation 3.6.2">(3.6.2)</a>), there exist constants <span class="process-math">\(c_1\)</span> and <span class="process-math">\(c_2\)</span> such that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_11.html ./knowl/eq3_14.html ./knowl/eq3_12.html ./knowl/eq3_12.html ./knowl/eq3_14.html">
\begin{equation*}
Y_1(x)-Y(x)=c_1 y_1+c_2 y_2 \to Y_1(x)=c_1 y_1+c_2 y_2+Y(x),
\end{equation*}
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<p class="continuation">which indicates <span class="process-math">\(Y_1(x)\)</span> is included in (<a href="" class="xref" data-knowl="./knowl/eq3_14.html" title="Equation 3.6.4">(3.6.4)</a>).</p>
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